Paper Physics CS(Phy) · (b)The soup in (a) is taken from a metallic container of heat capacity 2000 J q–1Ccontaining 16kgof soup maintained at 96 qC by an immersion heater. (i)

2020 HKDSE Physics & Combined Science (Physics)

Report on Assessment

Y.T.SZETOManager (Physics), HKEAA

7 & 8 Oct 2020

1

Overview

Paper Physics CS(Phy)

1A (MC)Mean: 17.5 out of 33

(i.e. 53%)(2019: 19.2 out of 33)

Mean: 8.9 out of 22 (i.e. 40%)

(2019: 9.8 out of 22)

1B ~50% (2019: ~>50%)

~35%(2019: 35%~40%)

2 ~<50% (2019: ~>50%) N.A.

SBA ~>70% (~2019) ~<70% (~2019)

Candidature ALL: 10692SCH: 9 866

ALL: 272SCH: 252

2

Marking & Grading

On-Screen Marking (OSM) panelsPhysics CS(Phy)

1B-1: Q.1, 3, 4, 5 (32M)1B-2: Q.6, 7, 8 (26M)1B-3: Q.2, 9, 10 (26M)

1B-1: Q.1, 2, 3, 4 (31M)1B-2: Q.5, 6, 7 (25M)

---2A: Astronomy (18%)2B: Atomic World (67%)2C: Energy (85%)2D: Medical Physics (30%)

---

SBA marks stat. moderated (outlying cases ~10% reviewed by Supervisors) 3

Marking & GradingExpert Panel (Examiners, 4 ~ 5 persons) determine level boundaries/cut scores based on Level descriptors / Group Ability Indicator (GAI) / Viewing candidate samples.

CS(Phy) graded by Common items / Viewing candidate samples.

Endorsement by Senior Management/Public Exam Board

Note: GAI is generated from Physics candidates’ actual percentage awards in 4 core subjects CEML taken into consideration the correlation between Physics and CEML.

4

1

Results

Level 5** 5+ 4+ 3+ 2+ 1+

Percentage 2.6% 26.3% 48.9% 72.6% 90.1% 97.9%

Level 5** 5+ 4+ 3+ 2+ 1+

Percentage 0.5% 5.0% 10.9% 35.8% 68.3% 88.1%

5

Physics

CS(Phy)

Cut score difference = 47.4 %

Cut score difference = 42.4 %

No. of MC 30 23/22 18 14/13 10 7

No. of MC 17 15 14/13 10 8 6

Paper 1APhysics (33 MC)

>70% 50%-70% <50% 4 16 13

E a s y D i f f i c u l t

CS (Phy) (22 MC)>70% 50%-70% <50%

2 3 17E a s y D i f f i c u l t 6

PHYSICS MC

Topic (No. of Qu.)Average

% correctNo. of Qu.

< 50% correct

Heat & Gases (3) 58% 0

Force & Motion (7) 59% 1

Wave Motion (11) 52% 6Electricity &

Magnetism (8) 48% 4

Radioactivity (4) 51% 27

CS(PHY) MC

Topic (No. of Qu.)Average

% correctNo. of Qu.

< 50% correct

Heat & Gases (2) 48% 1

Force & Motion (4) 47% 3

Wave Motion (9) 40% 7Electricity &

Magnetism (7) 36% 6

8

2

9

favourable distractor

5.

10


7.

11


12


3

13


14


15


16


4

17


ObservationsStrength:

- competent in calculations- most simple questions answered well

Weakness:- weak in handling units- qualitative responses far from precise- not paying attention to details when drawing

diagrams- weak in handling experimental type questions

18

Points to note

Paper 1: ~70% core partMethod marks ‘M’ and Answer marks ‘A’ (with tolerance range) adopted.Accept BOTH g = 9.81 or 10 m s 2.

19

Points to note

Electives (Total = 80 each) equating using Paper 1

Before equating: Mean 32 to 37 / SD 17 to 22After equating: Mean 36 to 42 / SD 17 to 19

2A Astronomy: ~ unchanged2B Atomic World: 2C Energy: 2D Medical Physics: ~ unchanged

20

5

Points to noteSamples of performance of candidates available in late October (HKEAA website).SBA cancelled for DSE Phy 20212021 DSE Phy Exam on 5 May 2021Markers’ Mtg: Paper 1B 11/5(tentative) Paper 2 10/5

21

THANK YOU

22

DSE PHYSICS 1B – Q 1, 3, 4, 5

Suggested Solutions Students’ Responses5 0.02 3300 (T – 4)

= 0.60 4200 (96 – T)

T = 85.347368 C 85.3 C

1. In a restaurant, ‘wontons in soup’ is prepared by putting 5 piecesof cooked wonton at 4 C into a bowl with 0.60 kg of soup attemperature 96 C.Given: average mass of each piece of wonton = 0.02 kg

specific heat capacity of wonton = 3300 J kg–1 C–1

specific heat capacity of soup = 4200 J kg–1 C–1

(a) Find the final temperature of the mixture. Assume that theheat capacity of the bowl and the heat loss to thesurroundings are negligible. (2 marks)

2

6

Students’ Responses

0

1

1

Missing ‘5’

Suggested Solutions


To compensate for / balance the heat loss (of the container with soup) to the surroundings

(b) The soup in (a) is taken from a metallic container of heat capacity2000 J C–1 containing 16 kg of soup maintained at 96 C by animmersion heater.(i) Why does that energy have to be supplied by the heater to

keep the soup at 96 C ? (1 mark)

0

1

1

Suggested Solutions Students’ ResponsesP 10 60

= 2000 9 + 16 4200 9

P = 1038 W 1040 W

1(b)(ii) A student used the following method to find the heater’s operating power P: remove the heater from the container and record the temperature of the 16 kg of soup after 10 minutes. It is found that the temperature has dropped 9 C. Estimate P.

(3 marks)

3

1

1

Suggested Solutions Students’ ResponsesSmaller than 9 C.

As the temperature of (the container with) soup drops, the rate of heat loss to the surroundings becomes lower due to a smaller temperature difference (w.r.t. the surroundings)

1(b)(iii) If the student repeats the measurement after another 10 minutes, would the corresponding temperature drop be larger than, equal to or smaller than 9 C ? Explain. (2 marks)

2

1

2

7

3. Read the following passage about a magnetically levitated(maglev) train and answer the questions that follow.

‘A maglev train car is just a box with magnets on the four corners,’ says Jesse Powell, the son of the maglev train inventor. The electromagnets employed have superconducting coils (i.e. coils with extremely low resistance). They therefore can generate magnetic fields 10 times stronger than ordinary electromagnets, enough to levitate and propel a train.

Two sets of magnetic fields are set up for different functions. One is to make the train float a few centimetres above magnetic poles P and Q as shown while the other is a propulsion system run by an alternating current for moving the train car along the guideway by magnetic attraction and repulsion. This floating design enables a smooth movement of the train. Even when the train travels up to 600 km per hour, passengers inside experience less vibration than travelling on traditional trains.

Suggested Solutions Students’ ResponsesWith superconductor / extremely low or nearly zero resistance,a much larger current flows in the coils producing a stronger magnetic field (with less heat loss due to the current flow).

(a) Explain why electromagnets employing superconducting coilscan produce much stronger magnetic fields. (2 marks)

train floats above magnetic poles P and Q

magnetic pole P

magnetic pole Q

NN

guideway

LEVITATION

2

1

0

Suggested Solutions


P and Q : N (north pole)

Repulsive force between like poles

(b) State the polarities of the magnetic poles P and Q andexplain how this arrangement enables the train to float. (2 marks)

2

1

10

Suggested Solutions Students’ Responses(i) As the train is not incontact with the rail,there is no friction /interaction between the train and the rail, thus smoother due to less vibration.(ii) There is no friction /interaction between thetrain and the rail, andthe propulsion of thetrain only needs to workagainst air resistance,therefore much faster

(c) Referring to the resistive forces experienced by the train, explainwhy a maglev train ride is (i) smoother and (ii) faster. (2 marks)

2

1

8

4. A small sphere is released from rest at point A and runs along asmooth track ABC as shown in Figure 4.1. The track around thelowest point B is approximately circular in shape.

The sphere leaves the track at point C where the track makes an angle of 60 with the horizontal. It finally reaches point E on the ground. Neglect air resistance. (g = 9.81 m s 2)

Suggested Solutions


vB > vC > vD > vA

(a) Arrange the speeds of the sphere at points A, B, C and Drespectively in descending order. (1 mark)

1

1

11

*(b) On Figure 4.1, use arrows to indicate the acceleration of the sphere, if any, at point B and at point D respectively. (2 marks)

Suggested solution

aB

aD


1

2

9


1

0

Suggested Solutions


Gravitational potential energy is changed to kinetic energy from A to B, and some kinetic energy is changed back to gravitational potential energy from B to C.

(c) (i) Describe the energy conversion of the sphere when it goesalong the track ABC. (2 marks)

1

2

Suggested Solutions Students’ Responses

(c)(ii) Hence find the speed of the sphere at point C. (2 marks)

2

0

Suggested Solutions Students’ ResponsesHorizontal speed at C = 4.85 cos 60= 2.43 m s 1

Distance = 2.43 t = 2.55 m

t = 1.05 s

*(c)(iii) If the horizontal distance between points C and E is 2.55 m, calculate the time of flight of the sphere before reaching point E.

(3 marks)

3

0

3

10

5. A block P of mass 10 kg is kept stationary on a smooth incline bya horizontal force F as shown in Figure 5.1. The incline makes anangle of 30 with the horizontal. (g = 9.81 m s 2)

Suggested Solutions Students’ Responses

(a)(i) On Figure 5.1, indicate and label all other forces acting on P.(2 marks)

Normal reaction R

Weight W

2

2


21

1 1

Suggested Solutions Students’ ResponsesN cos 30 = W = Mg and N sin 30 = F

F = Mg tan 30 = 56.6 N

R = N = = 113 N

Or

N = W cos 30 + F sin 30

and W sin 30 = F cos 30

(a)(ii) Find the magnitudes of the force F and the force exerted bythe block on the incline respectively. (3 marks)

3

3

11


1

0

Suggested Solutions Students’ Responsesg sin= 9.81 sin 30= 4.91 m s 2

(b) Now F is removed and neglect air resistance.

(i) What is the magnitude of the acceleration of the block ?(1 mark)

1

0

Suggested Solutions Students’ ResponsesDecrease

as the component of Fperpendicular to the incline no longer acts on the block / inclineOras F is removed, the force pressing the block / incline would decrease (only the weight’s component left).

(b)(ii) Explain whether the force exerted by the block on the incline would increase, decrease or remain unchanged when compared with that in (a)(ii). (2 marks)

2

2

0


1

0

0

12

22020 HKDSE - Physics1B-2

QUESTIONS 6, 7 & 8

6(a) You are given a hammer and a metal plate. Use ‘X’ to indicate a suitable location on Figure 6.1 where the hammer should hit the plate so as to generate a sharp loud sound to be received by the microphones in this experiment. State an additional piece of apparatus needed and the measurements to be made in this experiment.

X

Accept : ruler

6 (b)(i) Find the speed of sound in air. Show how you would treat the data obtained in the calculation.

(b)(ii) Suggest one adjustment to the experimental setting so as to obtain a more accurate result.

13

7(a)(i) Find iA.(ii) If iA is just greater than the critical angle of that boundary, estimate .

(iii) What phenomenon occurs at point A ? State the condition needs to be satisfiedby θ such that this phenomenon ffails to occur.

Well answered

7(b)(i) Explain why the light pulse detected is bbroader (i.e. of a longer duration) and with llower intensity. Assume that the loss of energy of the light pulse due to absorption by glass is negligible.

7(b)(ii) An engineer suggests changing the refractive index nc of the cladding in order to reduce the width of the light pulse received. should nc be increased or decreased ? Or, will a change in nc have no effect on the pulse width ? Explain your choice.

8(a) Explain why L1 and L2 are nnot lit.(b)(i) What is the potential difference across the taillight T2 ?

Accept : p.d. across each low-beam headlight is 0.

Well answered

14

8(b)(ii) Indicate on Figure 8.1 the direction of current in each of the branches AB, GB and BC. Which branch carries the largest current ?

8(c) Calculate the power delivered by the battery and show that the equivalent resistance of the circuit is slightly less than 1 in this setting.

8(d) Based on your answer in (c), explain whether a fuse rating of 15 A is suitable for this circuit or not.

Thanks

15

Candidates’ response - 6(a)

0A

1A0A

Candidates’ response - 6(b)(i)

1M0A

0M0A

Candidates’ response - 6(b)(ii)

0A

0A

Candidates’ response - 7(a)(ii)

0M0A

1M0A

1M1A

16

Candidates’ response - 7(a)(iii)

0A0A

0A1A

0A0A

1A1A

Candidates’ response - 7(b)(i)

Ignored

0A1A


1A0A

0A0A

Candidates’ response - 8(a)

1A

0A

0A

17


1A0A

0A0A

Candidates’ response - 8(c)

0M0A1M1A

0M0A1M0A

C

Answer from incorrect method

Candidates’ response - 8(d)

1M0A

0M0A

REVIEW OF 2020 HKDSE PHYSICS

EXAMPAPER 1B

Q. 2, 9 &10

18

QUESTION 2(a)(i)

Marking Scheme Candidates’ Performance

Well answered.

Some candidates did not indicate where the force acts.

Accept force without label [1A]

QUESTION 2(a)(i) Samples

Not showing where the force acts(not in contact with the piston) [0A]

QUESTION 2(a)(ii)


Quite a number of them mistook ‘force = pressure /area’.

Some calculated the force by the steam or the force due to atmospheric pressure instead of the force keeping the piston stationary.

(

Force = p/area [0M]Wrong answer [0A]

QUESTION 2(a)(ii) Samples

Fp = force by the steam [1M]= (3.10 106) 0.67

Wrong answer [0A]

Fp = force against atmospheric pressure [0M]Wrong answer [0A]

19

QUESTION 2(a)(iii)


A few took the temperature of steam to be 273 C or the Celsius temperature plus 237.

Some wrongly took the number of mole n to be 1.

pnRTV

61010.3)273237)(31.8)(018.0/570(V

QUESTION 2(a)(iii) Samples

Correct n [1M]Correct equation (correct values of P,

n and R, accept T = 237) [1M]Wrong answer [0A]

Wrong n [0M]Wrong equation (wrong T) [0M]Wrong answer [0A]

Correct n [1M]Wrong equation (wrong T) [0M]Wrong answer [0A]

QUESTION 2(a)(iii) Samples

Wrong n [0M]Wrong equation (R not

substituted & wrong answer) [0M]Wrong answer [0A]

QUESTION 2(b)

20

QUESTION 2(b)(i)


Most managed to find the work done by considering the gain of kinetic energy of the jet fighter.

Many candidates, who employed work done equals force distance, wrongly used the initial force and/or mistook the distance travelled as the product of the final velocity and time taken.

Weaker ones only found the net force instead of the work done or considered the work done as mass acceleration or force velocity.

QUESTION 2(b)(i) Samples

Wrong equation (F = calculated Fmax) [0M]

Wrong answer [0A]

Wrong equation (F = Fmax , d = vmaxx t) [0M]

Wrong answer (value & no unit) [0A]

Wrong equation (d = vmaxx t) [0M]

Wrong answer (value & unit) [0A]


Wrong equation (WD = power) [0M]

Wrong answer [0A]

Wrong equation (WD = average force) [0M]


No equation for WD [0M]WD not calculated [0A]

QUESTION 2(b)(ii)


Candidates did well in this part.

Some struggled to obtain ‘an average value’ for the correctly calculated average acceleration !

tuv

5.1054

21

Wrong equation(equation for initial acceleration) [0M]

Wrong answer [0A]

QUESTION 2(b)(ii) Samples

Wrong equation(wrong u & v, s = vmax t) [0M]

Wrong answer [0A]

Wrong equation (average acceleration =

‘acceleration’/time) [0M]Wrong answer (value & unit) [0A]

QUESTION 2(b)(iii)


Most candidates knew that the acceleration was decreasing but not many were able to give a clear explanation.

Some confused ‘decreasing’ with ‘decreased’ and stated that ‘friction reduces the speed and hence acceleration is decreasing’.

QUESTION 2(b)(iii) Samples

Correct answer [1A]No explanation for

decreasing pressure [0A]Decreasing force [1A]

Correct answer [1A]No explanation for

increasing air resistance [0A]Decreasing force [1A]

Correct answer [1A]No increasing resistance [0A]Force reduced

(not decreasing) [0A]


Correct answer [1A]Wrong explanation (speed

increases at a decreasingrate NOT decreases) [0A][0A]

Wrong answer [0A]Wrong explanation

(accelerating accelerationincreasing) [0A] [0A]

22

QUESTION 9(a) QUESTION 9(a)


Although most candidates were able to show the correct direction of the electric field, many failed to draw evenly spaced, parallel field lines properly.

QUESTION 9(a) Samples

Correct direction [1A]Perpendicular to

plates, paralleland evenlyspaced (Accept1 or 2 slightmistakes) [1A]

Correct direction [1A]Unevenly spaced [0A]

Correct direction [1A]Evenly spaced not

shown (only 2lines) [0A]

Correct direction [1A]Not perpendicular to

plates, not paralleland unevenlyspaced [0A]

QUESTION 9(a) Samples

Wrong direction [0A]Wrong pattern [0A]

Wrong direction [0A]Perpendicular to

plates, paralleland evenlyspaced [1A]

23

QUESTION 9(b)(i)


Not many were able to use the correct component of force or moment arm to find the moment acting on the rod.

Many mistook the length of the rod as the moment arm.

A few considered the moment to be the product of force and the separation of the plates.

Some mistook J or N as the unit of moment.


Correct equation (accept d as moment arm or wrongcomponent of F) [1M]


Correct equation [1M] Wrong answer

(value & unit) [0A]


Wrong equation (moment arm = separationbetween plates) [0M]


Correct equation (accept no component of F &moment due to force on onecharge) [1M]

Wrong answer [ 0A]

QUESTION 9(b)(ii)


Most candidates answered correctly.

Wrong units for the electric field were found occasionally.

dV

1.0100.5 3

24


Correct equation [1M]

Wrong answer (wrong or no unit) [0A]


Wrong equation [0M]Wrong answer

(value & unit) [0A]


(value & unit) [0A]

Correct equation (accept wrong order ofmagnitude of d) [1M]

Wrong answer [0A]


QUESTION 9(b)(iii)

Many candidates wrongly employed equations of force or electric field due to a point charge to tackle (b)(iii) 4

5

100.5100.2

EFQ

QF

QF


Wrong equation (wrong force) [0M]

Wrong answer [0A]


(value & unit) [0A]

25



(value & unit) [0A]

QUESTION 10(a)


Well answered.

Some candidates were unable to identify the product X despite they had found the correct proton and mass numbers.

QUESTION 10(a)

Accept ‘proton’ with wrong symbol [1A]

Accept ‘hydrogen atom’ [1A]

‘Isotope of H’ or ‘H2‘ not accepted [0A]

Correct atomic no. & mass no. without correct name or symbol not accepted [0A]

Wrong answer [0A]

QUESTION 10(b)


Most managed to find the mass defect correctly.

Some failed to convert the mass defect to energy in MeV.

Weaker ones mistook the energy equivalent of the mass of particle as the minimum kinetic energy or omitted the particle or the proton in finding the minimum kinetic energy.

26

QUESTION 10(b) Samples

Correct m [1M]Wrong answer (wrong method) [0A]

Correct m [1M]Answer not converted to MeV [0A]

Wrong m(mass of particle only) [0M]

Wrong answer [0A]

QUESTION 10(b) Samples

Wrong m(mass of omitted) [0M]

Wrong answer [0A]

Wrong m(mass of proton omitted) [0M]

Wrong answer [0A]

QUESTION 10(c)


Candidates’ performance was poor.

Only a few were able to give a correct explanation based on the conservation of momentum.

Popular explanations included ‘extra energy is needed because of inelastic collision or binding energy’, ‘to compensate for the energy lost to surroundings’ or ‘to overcome repulsive forces’.

QUESTION 10(c) Samples

Product(s) has/havemomentum [1A]

KE of product(s)not mentioned [0A]

Law of conservation of momentum only [0A]

KE of product(s) not mentioned [0A]

Inelastic collision [0A]KE of product(s) not

mentioned [0A]

Energy to ionize air molecules [0A]


27

QUESTION 10(c) Samples

Energy to breakthe bonds [0A]


Energy to overcomerepulsive force [0A]


THANK YOU!

Paper 2

Section A: Astronomy and Space Science

Q.1 Multiple-choice questionsA B C D

1.1 21.5 4.9 11.8 60.3*1.2 26.6 16.9 30.7 23.7*1.3 13.8 13.9 62.9* 7.41.4 14.2 44.1* 15.7 24.11.5 13.2 16.5 54.5* 12.91.6 9.0 56.5* 5.7 27.21.7 23.7 15.9 45.2* 13.01.8 51.7* 9.6 25.0 11.9

* : key ; Red colour : most favourable distractor

28

MCQ 1.2Two satellites X and Y are revolving in clockwise direction about the Earth as shown. The diameter of the circular orbit of X equals the length of the major axis of the elliptical orbit of Y. The two orbits intersect at P and Q.

P

Earth

Q

X Y

At the instant shown, the two satellites and the Earth are on a straight line. Which deductions are correct ?(1) X and Y have the same acceleration when they

pass P.(2) The speed of X is greater than that of Y at the instant

shown.(3) The satellites will not meet each other at P or at Q.

A. (1) and (2) onlyB. (1) and (3) onlyC. (2) and (3) only favourable distractor 30.9%*D. (1), (2) and (3 ) 23.7%

• At P, two satellites are at equal distance from the Earth.• According to Law of Gravitation, the acceleration of X and Y is equal at P (applying

also Newton’s 2nd law)• (1) is correct

• As the diameter of the circular orbit equals the major axis of the elliptical orbit,the circumference is larger for the circular orbit.

• Since the period is the same, average speed (circular) > average speed (elliptical)which is also greater than its lowest speed vY along the orbit (by Kepler’s 2nd law)

• (2) is correct

• According to Kepler’s 3rd law, the period of Y is the same as that of X.• Two satellites never meet at P or Q• (3) is correct

MCQ 1.4The luminosity of star P is double that of star Q. The brightness of P is eight times that of Q. What can be deduced about the respective distance of stars P and Q from the Earth ?

A. P’s distance from the Earth is 2 times that of Q.*B. Q’s distance from the Earth is 2 times that of P. 44.1%

C. P’s distance from the Earth is 4 times that of Q.D. Q’s distance from the Earth is 4 times that of P. 24.1%D. Q’s distance from the Earth is 4 times that of P. favourable distractor

8 = = 2L= total powerBrightness = 2

MCQ 1.7A star, which is 4.2 light years from the Sun, is observed from the Earth 6 months apart. Estimate the maximum angular difference in the observed positions of this star.

A. 0.8 arc seconds favourable distractor 23.7%B. 1.3 arc seconds*C. 1.6 arc seconds 45.2%D. 2.6 arc seconds

1 pc = 3.26 ly

i.e. 1.6 arc seconds after 6 months

= pc = 1.29 pc

= 0.8”

29

Q.1 Structured questionA certain star X at a distance of about 50 kpc from the Earth exploded a very long time ago and became Supernova 1987A (SN 1987A). The light from this supernova first reached the Earth in 1987.(a) Estimate how long ago the above explosion took place. (Give your answer in

years to 3 sig. fig.) (1 mark)

• Distance 50 kpc = 50000 3.26 ly = 163000 ly• Thus the explosion of the star took place 163000 years ago 1M• Note: 2020 – 1987 = 33 years can be ignored.

Most candidates knew how to find when the explosion occurred although some of them made mistakes in units conversion.

(b) At maximum brightness, SN 1987A has an apparent magnitude of +2.9. Is theabsolute magnitude of SN 1987A at maximum brightness smaller than, larger thanor equal to +2.9 ? Explain your answer. (2 marks)

SN 1987A at 50 kpc, being much farther away than 10 pc, would be muchbrighter (than that corresponding to +2.9) if it were placed at 10 pc. 1A

No mark awarded without ‘placed at 10 pc’Hence (the numerical value of) its absolute magnitude is much smallerthan +2.9 / the apparent magnitude. 1A

Some candidates failed to point out that the absolute magnitude corresponds to observing the star as if it were placed at 10 pc.

30

Before the explosion of star X, its luminosity was about 40000 times that of the Sun and its surface temperature was 3.1 times that of the Sun.

(c)(i) Use Stefan’s law to show that the radius of the star X is about 20 times the Sun’s radius. (2 marks)

Correct expression derived from Stefan’s law and

1M

Correct sub. into the correct expression

1M

Well answered. A few candidates forgot to start with Stefan’s law.

4

S

X2

S

X

S

XTT

RR

LL

42

S

X 3.140000RR

8.2081165.20S

XRR

Did not start with Stefan’s law

Without correct ratio

(c)(ii) In which region, A, B, C or D, on the Hertzsprung-Russell diagram was star Xlocated ? Explain whether or not this star can be classified as a ‘red giant’.

(2 marks)

0.0001

0.001

0.01

0.1

1

10

100

1000

10000

100000

1000000

100010000

lum

inos

ity(s

olar

lum

inos

ity)

surface temperature (K)

Sun

region A region B

region C region D

Region A. 1ANot a ‘red giant’ as the temperature of star X is (much) higher than that of the Sun. / 1ARed giants are in region B

Well answered. Most were able to identify features of star X that ruled out the possibility of classifying it as a ‘red giant’.

31

(d) A special feature of SN 1987A is that a circular ring of gas surrounds the supernova.The gas was ejected by the star X some time before it exploded. As shown inFigure 1.1, each point on this ring is expanding outwards at a constant speed fromthe supernova.

Top view

distant observer

P

Qexpanding ring of gas

supernova

R

S red violetL1 L2

L0

spectral line from the ring of gas

Suppose a distant observer on the plane containing the ring views a certain spectral line from the ring of gas and finds that it covers wavelengths between the limits L1and L2 as shown in Figure 1.1. L0 is the wavelength of that spectral line when observed in the laboratory. State the respective wavelengths that originate from point Q and point R on the ring. Explain your answer. (3 marks)

Q: L0 or R: L1

According to Doppler effect, gas at R receding from the observer gives rise to red shift (vice versa for P)Accept: Moving away from the observer

1A

1A

1AWhile there is no velocity component for gas at Q (and S) towards / away from the observer, no Doppler / blue / red shiftAccept: No velocity along the direction toward the observer.

The velocity of Q is perpendicular to the line joining Q and the observer

NOT accept: no relative velocity between Q and observer

Some candidates had a misconception that there was no Doppler effect as the distance of the star from the observer did not change.

Paper 2

Section C : Energy and Use of Energy

32

Q.3 Multiple-choice questions

A B C D3.1 12.2 65.3* 16.1 6.33.2 83.0* 6.8 7.3 2.83.3 7.2 11.6 8.2 72.9*3.4 13.1 8.5 54.6* 23.73.5 65.0* 6.1 11.1 17.93.6 5.6 24.1* 56.5 13.53.7 8.2 30.9 13.2 47.7*3.8 85.3* 11.0 0.7 3.0

* : key ; Red colour : most favourable distractor

MCQ 3.6

24.1%favourable distractor 56.5%

= =

=

=

= -

=

’=

*

MCQ 3.7

47.7%

30.9%favourable distractor

condenser evaporator

At expansion valve, liquid refrigerant expands.

Compressor compresses vaporized refrigerant to a high temperature and high pressure.

*

Q.3 Structured question

33

In (a), not many were able to mention that an atom or a nucleus is more stable if it possesses a higher binding energy per nucleon.

1M

1M

0MQ.3 Structured question

0M

0MIn (b), many candidates did not fully understand what ‘binding energy’ means and failed to state its physical meaning of ‘energy required to separate the nucleons completely’. Weaker ones held that it was the amount of energy needed to ‘bind’ the nucleus from its constituent nucleons.

For the calculation in (b)(ii), some incorrectly multiplied the binding energy per nucleon by the difference of mass number and atomic number.

0M

0M


34

In (c)(ii), few were able to mention that chain reaction could not be sustained when the concentration of U-235 in the fuel rods was too low.

1M 0M

Less than half of the candidates obtained the correct answer in (c)(i). Many made mistakes in units conversion or in manipulating the efficiency of energy conversion.

1M0M

0M


In (d), most candidates knew the function of moderator in a fission reactor though a few wrongly stated that ‘electrons’ or ‘atoms’ were to be slowed down.For those who pointed out that control rods were to absorb neutrons, some did not further explain that they were for controlling the rate of reaction.

1M

1M

0M

Paper 2

Section B: Atomic World

HKDSE 2020

35

Multiple Choice

Qn. 1 2 3 4 5 6 7 8

A 38.9% 4.0% 29.4% 9.1% 40.2% 42.4% 11.2% 8.1%

B 16.7% 6.0% 26.5% 42.1% 19.8% 28.6% 61.3% 43.6%

C 28.4% 74.5% 9.5% 25.0% 21.3% 15.8% 12.6% 15.5%

D 15.2% 14.6% 33.6% 22.9% 17.9% 12.1% 14.7% 32.0%

KEY: underlined

Qn. 2.1

Answer : C (28.4%)Best distractor: A (38.9%)Many candidates did not understand alpha particles being more massive is thekey of the theory behind scattering experiment. They therefore failed to applythe concept to an unfamiliar case of beta particles.

X

W Z


Qn. 2.3

X

Y

W Z

Answer : B (26.5%)Best distractor: D (33.6%), A (29.4%)Not many candidates were able to find the correct wavelength corresponding to an energy change of 13.6 eV. It is worth to note that 1/3 of the candidates thought that the spectra of n=1, n=2 etc. have common lines.


Qn. 2.4

Answer : D (22.9%)Best distractor: B (42.1%)Many candidates did not know that Bohr model can be applied to any single-electron atom or ion. They were not familiar with the reason behind theformula .


36

Qn. 2.8

Answer : D (32.0%)Best distractor: B (43.6%)Many candidates did not know melting point depression effect ofnanoscale materials.


Q.2 Structural question


In (a)(ii), most candidates misunderstood the question. Instead of giving the implication of the experimental result, they tried to explain why there is such a result. Some candidates’ explanation was not precise or not relevant, such as one-to-one process.


For (b)(i), some candidates also misunderstood the question. They did not answer the question based on the voltage = 1.7 V. Many said that the current increases as there are more photons, i.e. the situation before the voltage is set to 1.7 V.

37

Q.2 Structural Question

Performance of (b)(ii) is satisfactory. Common mistakes:• some candidates simply took the photon energy hf as the work

function.• some candidates had difficulties in conversion between joule and eV.


Performance in (c)(i) is fair. Some candidates did not realise the number of photoelectrons can be directly found from the current given. They tried to find the answer by photon energy, work function, etc., which is indeed not possible.

Q.2 Structural Question

For first 1A, many candidates did not have a solid concept about the meaning of different energies. Even though the answer can be obtained simply by subtracting the two values given, they cannot figure out the correct way to find the answer. For second 1A:• Some candidates thought the variation is caused by things happen

when the photoelectrons move from cathode to anode (rather than theprocess when electrons escape from the metal surface).

• Some other candidates employed the concept in Bohr model ofhydrogen atom to explain what happens in metal.

HKDSE 2020Physics Paper 2

Section D: Medical Physics

38

Multiple Choice Questions

Qn 1 2 3 4 5 6 7 8

A 9% 13% 5% 46% 16% 15% 24% 37%

B 20% 14% 28% 13% 43% 3% 55% 36%

C 12% 59% 8% 27% 29% 76% 5% 13%

D 59% 14% 59% 14% 12% 6% 16% 14%

14

KEY: underlined

Qn. 4.4Which of the following statements about the piezoelectric crystal inside an ultrasound transducer is/are correct ?(1) The piezoelectric crystal converts electrical signals into

mechanical vibrations and vice versa.(2) The thickness of the piezoelectric crystal is arbitrary.

X

YA.* 46% Only (1) is correct.B. 13% Only (2) is correct.C. 27% (1) and (2) are correct.D. 14% (1) and (2) are incorrect.

15


Qn. 4.5Using the information given below, find the proportion ofenergy transmitted when ultrasound is incident from air toskin.

X

Y

W Z

A. 16% 5.7 × 10-4

B.* 43% 1.1 × 10-3

C. 29% 2.8 × 10-3

D. 12% 1.0 ×10-2

16


acoustic impedance / kg m 2 s 1

air 430

soft tissue 1.5 ×106

Qn. 4.8The figure shows a simplified diagram of the cross-section ofhuman arm. The medullary cavity is the central cavity that isfilled with bone marrow. The linear attenuation coefficient ofbone marrow is roughly the same as that of muscle.

Y

17

muscle

bone

medullary cavity (filled with bone marrow)

X-rays

39

Qn. 4.8Which figure below best represents the X-ray radiographicimage of the arm ?

Y

A.*

37%

B.

36%

C.

13%

D.

14%

18


Q.4(a)(a) State briefly how X-rays are produced.

19

Q.4(a) Q.4(b)(i)(b)The table below shows the linear attenuation coefficients of soft tissue

and bone for an X-ray beam.

21

linear attenuation coefficient

soft tissue s = 0.51 cm-1

bone b = 2.46 cm-1

(i) State one factor contributing to a higher linear attenuationcoefficient of bone compared to soft tissue. (1 mark)

40

Q.4(b)(i) Q.4(b)(ii)(b)The table below shows the linear attenuation coefficients of soft

tissue and bone for an X-ray beam.

23

linear attenuation coefficient

soft tissue s = 0.51 cm-1

bone b = 2.46 cm-1

(ii) X-ray beam of intensity I0 passes through soft tissue that is5.6 cm in thickness and is attenuated to intensity I. Whatthickness of bone would yield the same degree of attenuationof the same X-ray beam ? Show your calculation. (2 marks)

Q.4(b)(ii) Q.4(b)(ii)

41

Q.4(b)(iii)(ii) Explain why X-ray radiographic imaging of the breast usually employs X-

rays of lower energy (~20 keV) while examination of bony structuresemploys X-rays of energy around 100 keV. (2 marks)

26

Q.4(b)(iii)

Q.4(b)(iii) Q.4(b)(iii)

42

Q.4 (c)People are often concerned about the radiation exposure during medical examinations like X-ray radiographic imaging and computed tomography (CT) scans. Some information about the radiation dose is given below:

30

Source / item Equivalent dose

Taking an X-ray radiographic image 0.1 – 0.2 mSv

A CT scan 1 – 10 mSvAverage weekly natural background dose for a person about 0.05 mSv

Q.4(c)(i) State one potential hazard of exposure to ionizing radiation on

human bodies. (1 mark)

(ii) Explain why the equivalent dose of a CT scan is higher than thatof taking an X-ray radiographic image. (2 marks)

(iii) Name one source contributing to the natural background dose.(1 mark)

31

Q.4(c)(i) Q.4(c)(i)

43

Q.4(c)(ii)(ii) Explain why the equivalent dose of a CT scan is higher than that

of taking an X-ray radiographic image. (2 marks)

34

Q.4(c)(ii)

Q.4(c)(iii)(iii) Name one source contributing to the natural background dose.

(1 mark)

36

Q.4(c)(iii)

44

The End

45

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